The figure shows rectangle $ABCD$ with segment $PQ$ dividing the rectangle into two congruent squares. How many right triangles can be drawn using three of the points $\{A,P,B,C,Q,D\}$ as vertices? [asy]
draw((0,0)--(8,0)--(8,4)--(0,4)--cycle);
draw((4,0)--(4,4));
label("D",(0,0),S);
label("Q",(4,0),S);
label("C",(8,0),S);
label("B",(8,4),N);
label("P",(4,4),N);
label("A",(0,4),N);
[/asy]
Explanation: First, we consider the triangles that have the vertices of the rectangle $ABCD$ as the right angle. We can get $2$ right triangles for each vertex. For example, for vertex $A$, we can get right triangles $DAP$ and $DAB$. Since there are four vertices, we can get $2 \cdot 4 =8$ right triangles.

Next, we consider triangles that have $P$ or $Q$ as the vertices. We can set $PQ$ as a leg of the right triangles and get $4$ right triangles with the third vertex $A,B,C$, and $D$.

Lastly, we can draw the diagonals $DP, CP, AQ$, and $BQ$. Since $ADQP$ and $BCQP$ are squares, each diagonal creates a $45$ degree angle with the line segment $PQ$. Therefore, we have two right triangles: $DPC$ and $AQB$.

Adding them together, we have a total of $$8+4+2=\boxed{14 \text{ right triangles}}.$$